How Is the Electric Field Intensity Calculated on a Charged Sphere's Surface?

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Homework Help Overview

The discussion revolves around calculating the electric field intensity on the surface of a charged sphere, given its electric potential. The problem involves understanding the relationship between electric potential and electric field strength, particularly in the context of a sphere with a specified radius and charge.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the derivation of the electric field formula and question the presence of squares in the equations presented. There are attempts to clarify the relationship between voltage and electric field strength, as well as the dimensions involved in the formulas.

Discussion Status

The discussion is ongoing, with participants expressing confusion about the formulas and seeking clarification on the derivation of the electric field intensity. Some guidance has been offered regarding the dimensions of the quantities involved, and there is an acknowledgment of the need to verify the correctness of the equations used.

Contextual Notes

Participants note discrepancies in the formulas provided and question the accuracy of the original poster's equations. There is a focus on ensuring that the mathematical relationships are correctly understood, particularly in the context of homework constraints.

  • #31
And when we flash back to post #4 ...
 
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  • #32
That's flashing back even further !

But the same link has something to say about the derivation too, I seem to remember. The name Gauss comes to mind !
I take it you can look around there all on your own (with a little patience, perhaps ...) , so -- like a good helpdesk worker -- I can now safely close this ticket, I hope ?
 
  • #33
ehild said:
Do you remember Coulomb's Law?
##F##=##\frac{K Qq}{r^2}##
 
  • #34
gracy said:
I am still clueless about my question in op.
##E##=##\frac{\left(\frac{1}{4πε0}\frac{q}{R}\right)^2}{\frac{q}{4πε0}}##
They took the equation ##E##=##\frac{1}{4πε_0}\frac{q}{R^2}## and rewrote it in that form. You can check they are equivalent. This was useful because the numerator is the same as V2 and the denominator only involves the charge and some constants. Since V and q are known, this gives a quick way of finding E.
 
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